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Chapter 7
The Early Universe: Neutrinos,
Nucleosynthesis and
Recombination
With telescopes we can look all the way back to the time when the CMB was released,
at z = 1100, i.e. 0.3 million years after the Big Bang. Unfortunately the CMB blocks
our view of what happened earlier. We have to infer the cosmic line of events before
the CMB using theoretical modeling. Fortunately there are some pieces of evidence
today that are relics from the processes that happened during this early stage of the
Universe, the most evident being (a) the primordial abundances of the elements and
(b) the anisotropies in the CMB and structures that formed in the Universe lateron. We
will discuss (b) in Chapter 5. In the present chapter we will discuss the thermal processes that happened in the early Universe and how they created the initial abundances
of the elements in our Universe.
7.1
Assumptions
To a surprising degree of accuracy we can assume that the Universe behaves adiabatically, i.e. during the expansion we can assume that all processes are reversible. And it
also turns out that we can understand many of the thermal processes in the early universe with thermal equilibrium followed by “freeze-out”. What it means is that before
that freeze-out time the abundances of species of particles are given by their thermal
equilibrium value (which changes with time as the universe cools down), while after
that time the abundances stay fixed. This means that we can calculate the abundances
to good approximation by finding at which temperature they froze out and taking the
thermal equilibrium abundances at that temperature.
This assumption requires that the reactions that convert the different species of particles into each other and/or create/destroy particles are fast enough that at all times
before freeze-out the system is in thermal equilibrium. It turns out that this is a reasonable approximation.
So in this chapter we will study the thermal evolution of the early Universe as a series
of equilibrium states for all particles that have not yet frozen out.
7.2
A recap of some statistical physics principles
To derive the equilibrium abundances of particles we need some tools from statistical
physics. I assume that you are familiar with them, so we will not rigorously derive
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them from scratch. Some more in-depth discussion can be found in the script on
cosmology by Matthias Bartelmann.
7.2.1 Statistical equilibrium of quantum states of identical particles
Suppose we have N identical particles. Each can acquire energy states !m with m =
0, 1, 2, 3, · · · , M. A point in this phase space can be denoted as (m1 , m2 , · · · , mN ),
meaning that particle 1 is in state m1 (having energy !m1 ), particle 2 is in state m2 etc.
Each phase space point has many “twins”: other phase space points that are physically
the same. Example: (1, 0, 0) is the same as (0, 1, 0) etc. Let us call each point in
phase space a microstate and the the total set of identical microstates a distribution.
A nice way to denote a distribution is to count how many of the N particles are in
!
each one-particle energy state: {N0 , N1 , · · · , N M }, where m Nm = N. Example: The
distribution belonging to microstate (0, 0, 1, 1, 0, 2, 0) would be {4, 2, 1, 0, 0, · · ·}, but
also microstate (1, 1, 0, 0, 0, 2, 0) has the same distribution.
A fundamental principle of statistical physics is that all microstates have the same
probability, provided some constraint equations are fulfilled. The constraint equations
are, in our example, that the total number of particles is a given value (N) and that the
total energy also equals a given value (E). The distribution which corresponds to the
largest number of identical microstates is therefore the most likely distribution. The
result is (without proof):
N
Nm = e−!m /kT
(7.1)
Z1
where T is the temperature and k is the Bolztmann constant. The symbol Z1 is a
!
normalization constant such that m Nm = N and is called the partition sum:
Z1 =
M
"
e−!m /kT
(7.2)
m=0
Often quantum states are degenerate: they in fact consist of a multitude of states with
identical energy !, typically as a result of rotational symmetry of the particle. Rather
than treating each of these states separately, we pack them together as a single state m
with energy !m and a degeneracy gm (also called “statistical weight” gm ). We then get
Nm
=
Z1
=
Ngm −!m /kT
e
Z1
M
"
gm e−!m /kT
(7.3)
(7.4)
m=0
Later, when we will allow particles to be created, destroyed and/or transformed into
another type of particle (through e.g. chemical reactions), it will be important to include also the chemical potential µ to these equations. Without proof this leads to
Nm
=
Z1
=
Ngm −(!m −µ)/kT
e
Z1
M
"
gm e−(!m −µ)/kT
(7.5)
(7.6)
m=0
!
Note, however, that for a system of fixed N = m Nm the inclusion of µ in the above
equations does not change anything: µ drops out of Eq. (7.5) and thus Eqs. (7.5,7.6)
are equivalent to Eqs. (7.3,7.4).
7.2.2 Maxwell-Boltzmann distribution
If we consider the movement of the N particles, then their “quantum states” are their
momenta "p = m"v where m is the particle mass (not to be confused with the index m for
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the quantum states). In ("x, "p)-space one quantum unit has volume h3 . Eqs. (7.3,7.4)
should then be applied to these quantum units. Since the quantum units of ("x, "p)-space
are so small, the sum in Eq. (7.6) can be expressed as an integral:
#
#
gV
4πgV ∞ −(!(p)−µ)/kT 2
Z1 = 3
e−(!("p)−µ)/kT d3 p =
e
p dp
(7.7)
h
h3
0
The energy !(p) is, in the low-velocity-limit of special relativity, expressed as
p2
2m
!(p) = mc2 +
(7.8)
So Z1 becomes
#
4πgV −(mc2 −µ)/kT ∞ −p2 /(2mkT ) 2
e
e
p dp
h3
0
gV(2πmkT )3/2 −(mc2 −µ)/kT
e
=
h3
Z1 =
(7.9)
According to Eq. (7.5) the number of particles Nm per quantum state m is
Ng −p2 /(2mkT ) −(mc2 −µ)/kT
e
e
Z1
Nh3
2
e−p /(2mkT )
=
V(2πmkT )3/2
Nm =
(7.10)
Again: The m in Nm is an index, while the m in the rest is the particle mass.
Now it is convenient to define a one-particle distribution function f ("p) such that
f ("p)d3 p describes the chance of finding a given particle in a box$ of momentumvolume d 3 p around momentum vector "p. It therefore holds that f ("p)d3 p = 1.
In a spatial volume V a quantum unit in "p-space has ∆p3 = h3 /V, and thus Nm =
N f ("p)h3 /V. This gives
1
2
e−p /(2mkT )
(2πmkT )3/2
$∞
If we define f (p) = 4πp2 f ("p) (such that 0 f (p)d p = 1) we arrive at
f ("p) =
f (p) =
4π
2
p2 e−p /(2mkT )
(2πmkT )3/2
(7.11)
(7.12)
This is the Maxwell-Boltzmann distribution.
7.2.3 Chemical reactions in equilibrium
If we have multiple species in the same box, then each of these species has its own
chemical potential µi . Suppose we have a chemical reaction
α1 Species1 + α2 Species2 ↔ β3 Species3 + β4 Species4
(7.13)
We can define a parameter ξ that determines how far the reaction has proceeded (in
the direction from 1,2 to 3,4):
dN1
= −α1
dξ
dN2
= −α2
dξ
dN3
= β3
dξ
dN4
= β4
dξ
(7.14)
According to the rules of statistical physics (without proof) the equilibrium of this
reaction (with constant volume and temperature) is found when
dF(ξ)
=0
dξ
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(7.15)
where F is the free energy of the system (we will come back to this later). Eq. (7.15)
says that the system in chemical equilibrium (assuming only the above chemical reaction is possible) when we are at a minimum of F along the reaction track. This means
that
dF(ξ)
∂F
∂F
∂F
∂F
= −α1
− α2
+ β3
+ β4
=0
(7.16)
dξ
∂N1
∂N2
∂N3
∂N4
where the notation is such that
&
%
∂F
∂F
≡
∂Ni
∂Ni N j!i =const
(7.17)
If we define the chemical potentials µi as
%
&
∂F
µi =
∂Ni N j!i =const
(7.18)
then we get
− α1 µ1 − α2 µ2 + β3 µ3 + β4 µ4 = 0
(7.19)
We can also express dN2 , dN3 and dN4 all in terms of dN1 , because we have to move
along the chemical reaction:
dN2 =
α2
dN1
α1
dN3 = −
β3
dN1
α1
dN4 = −
β4
dN1
α1
(7.20)
In that case the dF(ξ)/dξ = 0 condition becomes
dF(N1 )
=0
dN1
(7.21)
where it is understood that this is along the reaction path. This way of writing is also
often useful, if we know the expression for F(N1 , N2 , N3 , N4 ).
It is essential in this theory to understand that the values of N1 , N2 , N3 and N4 in
equilibrium are found because of their constraint: any increase in N4 goes along with
a decrease in N1 .
There is also a special kind of reaction: Particle creation, like the creation of photons:
α1 Species1 + α2 Species1 ↔ α1 Species1 + α2 Species1 + βSpecies1
(7.22)
for instance:
p+e↔ p+e+γ
(7.23)
In this case one could say β1 = α1 and β2 = α2 so the equation dF(ξ)/dξ = 0 becomes
− α1 µ1 − α2 µ2 + β1 µ1 + β2 µ2 + β3 µ3 = β3 µ3 = 0
(7.24)
µ3 = 0
(7.25)
This means
The chemical potential of this particle 3 (in our example this is a photon) must therefore be 0. This is because the number of photons is not constrained by any constraint
equation.
We can also do a mixed-form reaction: partly constrained and partly unconstrained.
An example is ionization/recombination:
p+e↔ H+γ
(7.26)
The constraint is
dN p = dNe
dNH = −dN p
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(7.27)
In principle one would also have dNγ = dNH . In that case we would have α p = 1,
αe = 1, βH = 1 and βγ = 1, i.e. µ p + µe = µH + µγ . But often there is no constraint on
Nγ , for instance if one is assumed to be in a known radiation field where Nγ % NH , so
that an ionization event does not affect Nγ very much. In that case we keep µγ fixed.
µ p + µe = µH + µ(fixed)
γ
(7.28)
If, on top of that, the radiation field is thermal, then µγ = 0, in which case we obtain
(7.29)
µ p + µe = µH
7.2.4 Saha equation
Using the above rules we can derive the equilibrium values of Ne , N p and NH under the
presence of ionization/recombination reactions. Since we want to find the minimum
of the free energy F we express F first in terms of the canonical partition function Zc
of the combined system of three species (without proof):
F = −kT ln(Zc )
(7.30)
with (also without proof; see lectures on statistical physics):
N
Zc =
ZeNe Z p p ZHNH
Ne !N p !NH !
(7.31)
where Ze , Z p and ZH are the one-particle partition functions for electrons, protons and
hydrogen atoms, respectively (cf. Eq. 7.9).
Let us define the baryon number Nb = N p + NH which is conserved. The hydrogen
density can be expressed in terms of Nb and Ne by NH = Nb − Ne . For the ln(N!) terms
we use Stirling’s formula ln(N!) = N(ln N − 1). If we take as our reaction parameter
dξ = dNe we have
dN p
dNe
dNH
=1
=1
= −1
(7.32)
dξ
dξ
dξ
We can now express dF(ξ)/dξ as
0=
d'
dF(ξ)
Ne ln Ze + N p ln Z p + NH ln ZH
=
dξ
dξ
− Ne (ln Ne − 1) − N p (ln N p − 1) − NH (ln NH − 1)
= ln Ze + ln Z p − ln ZH − ln Ne − ln N p + ln NH
This immediately implies
Ne N p Ze Z p
=
NH
ZH
(
(7.33)
(7.34)
With Eq. (7.9), together with (me + m p − mH )c2 = 13.6 eV and µe + µ p − µH = 0 and
ge g p /gH = 1, and the identities Ne = N p and NH = NB − Ne we obtain
Ne2
V(2πme kT )3/2 −χ/kT
=
e
Nb − Ne
h3
(7.35)
with χ = 13.6 eV. Here we used m p /mH & 1. We can write this with x = Ne /Nb and
nb = Nb /V as
(2πme kT )3/2 −χ/kT
x2
=
e
(7.36)
1−x
h3 nb
This is Saha’s equation for thermal ionization equilibrium.
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7.2.5 Equilibrium number densities without constraints
If a particle can be created and destroyed without constraints (other than the energy of
course), we have already seen that their chemical potential µ in thermal equilibrium
is 0. This is, for instance, the case for photons, but at very high temperatures this
might also be approximately the case for other particles. For instance, at very high
temperatures you can create e+ e− pairs in abundance, as much as the available energy
allows you. As long as this number is much larger than the pre-existing electrons,
the abundance of e+ e− pairs is nearly unconstrained, and will have a near-zero chemical potential. This leads to a well-defined occupation number of each quantum state
(without proof):
1
Nm = !/kT
(7.37)
e
±1
where the + is for fermions and the − for bosons.
For photons with ! = hν we thus get
Nm(Planck) =
1
ehν/kT
(7.38)
−1
Since the density of quantum states (per volume per frequency) is ρ s = 4πgν2 /c3 (with
g = 2 for photons), and the energy per photon is hν we see that the equilibrium energy
for light U(ν) is
4πghν3/c3
(7.39)
U(ν) = hν/kT
e
−1
Now, per sterradian (dividing by 4π) and passing through a surface of 1 cm2 per second
(multiplying with c) this gives the Planck function:
2hν3 /c2
(7.40)
ehν/kT − 1
where we inserted g = 2. This is the intensity of thermal radiation of a given temperature T (blackbody radiation), and it is an extremely good approximation to the
intensity of the CMB if we take T = T CMB = 2.725 K.
Bν (T ) =
A similar exercise can be made for fermions such as electron/positron pairs, but we
will not do this here.
We can also compute the total number of particles of a certain kind in equilibrium,
!
which is N = m Nm . After some algebra (see, if you are interested, the script by
Bartelmann) one obtains
% &3
ζ(3) kT
3 gF
nB = gB 2
,
nF =
nB
(7.41)
!c
4 gB
π
where n = N/V, and F stands for fermion and B for boson. Here ζ(3) = 1.202 is the
Riemann Zeta-function evaluated at 3. This gives nB & 10.14 gB(T/K)3 cm−3 . Note
that this means that for the CMB at the present time, with T CMB = 2.725 K we get
about 400 photons per cm3 . Most of these photons are at energies close to the peak of
the Planck function.
Likewise we can calculate the total energy U for the particles. The result is:
uB = gB
π2 (kT )4
30 (!c)3
uF =
,
7 gF
uB
8 gB
(7.42)
where u = U/V. For photons this gives an average energy per photon of 'hν( = 2.7 kT .
The total entropy S is
sB = g B k
% &3
2π2 kT
45 !c
,
where s = S /V.
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sF =
7 gF
sB
8 gB
(7.43)
7.3
Thermal evolution of the early Universe
7.3.1 Adiabatic expansion of thermal radiation
We already saw in Section 4.4 that adiabatic expansion of an isotropic radiation field
leads to
1
1
ρ ∝ p ∝ 4 ∝ 4/3
(7.44)
a
V
where the volume V ∝ a3 . If this radiation field consists of photons in thermal equilibrium, then from Eq. (7.42) we find (with uF ≡ ρ)
T∝
1
1
∝ 1/3
a V
(7.45)
The temperature thus scales in the same way as the frequency of the photons ν ∝
1/a. This means that if we follow a photon during the expansion of the Universe,
the !m /kT in the expression for the occupation number of quantum states (Eq. 7.37)
stays the same. The occupation numbers Nm of the quantum states do not change: Just
the energy !m corresponding to those states goes as 1/a. This has a very important
consequence: The spectrum of a thermal radiation field remains thermal, even after
adiabatic expansion; only the temperature belonging to that thermal radiation field
decreases. This means that the CMB is a thermal spectrum today, both in terms of the
spectral shape as well as in terms of the photon density.
7.3.2 Adiabatic expansion of non-relativistic ideal gases
For non-relativistic matter we have already seen that, to first order, ρ ∝ 1/a3 . However,
this tells nothing about the temperature of that matter, since this relation holds in the
limit of infinitely cold matter. To find how the temperature of non-relativistic gas
behaves we must look at the adiabatic index γ of the gas, defined by
p ∝ ργ
(7.46)
The γ is related to the number of degrees of freedom n that the particles of the gas
have:
n+2
γ=
(7.47)
n
For point particles one has n = 3 because of the x, y and z directions of motion. This
gives γ = 5/3. For a diatomic molecule, for instance, we have two additional degrees
of freedom provided by rotation, i.e. for such gas we have n = 5 and thus γ = 7/5.
The more degrees of freedom the particles have the more γ approaches 1, i.e. the
“softer” the matter becomes. Note that in the notation of Eq. (7.46) radiation would
have γ = 4/3.
For an ideal gas consisting of particles with mass m one has
p=
ρkT
m
(7.48)
This means that as the Universe expands the temperature of this non-relativistic gas
goes as
p
T ∝ ∝ ργ−1 ∝ a3(1−γ)
(7.49)
ρ
For monoatomic gas (γ = 5/3) we thus have
T∝
1
a2
This means that non-relativistic gas cools faster than radiation.
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(7.50)
7.3.3 Particle freeze-out
Let us verify if the assumption that we have thermal equilibrium at large a followed
by freeze-out at smaller z is correct. We do this by setting up a reaction rate equation
for the production and destruction of some particle species, which we call s. Let us
call the number density of this species n s . Such a particle can be created by a collision
between two other particles, who’s number density we write as na and nb . We write
the collisional cross section between particle type a and particle type b as σab , and we
define pab→sc the probability that upon such a collision a particle of type s is formed,
including some additional particle c (for momentum and energy conservation). If the
average relative velocity between particles a and b is written as '∆vab (, then the rate
of production of particles of type s (and thus also of type c, but that is irrelevant here)
is:
jab→sc = na nb '∆vab (σab pab→sc
(7.51)
Now let us assume that a particle of type s can be destroyed again by collisions with
some particle of type c with number density nc (and thus create particles a and b
again).
j sc→ab = n s nc '∆v sc (σ sc p sc→ab
(7.52)
You see that jab→sc does not depend on n s , but j sc→ab does. Therefore it is convenient
to define
j :=
α :=
jab→sc = na nb '∆vab (σab pab→sc
j sc→ab /n s = nc '∆v sc (σ sc p sc→ab
(7.53)
(7.54)
so that neither j nor α depend on n s . The symbol α is the destruction rate per particle
of type s. The average life time of these particles is
τlife = 1/α
(7.55)
At any redshift z the change in the number density n s obeys the following equation:
ṅ s + 3Hn s = j − αn s
(7.56)
If both j = 0 and α = 0 (no collisions), then Eq. (7.56) reduces to
ṅ s = −3Hn s
(7.57)
which implies that n s ∝ 1/a3. This is simply the conservation equation for number
density in an expanding Universe.
If both j and α are extremely large, then Eq. (7.56) implies that n s is given by
ns =
j
α
(7.58)
which is the equilibrium value for n s . Let us denote this equilibrium value as n sT . We
can thus rewrite Eq. (7.56) as
&
%
n sT
−1
(7.59)
ṅ s + 3Hn s = n s α
ns
√
If we go back in time to very small a, then α increases rapidly: We have α ∝ na nb T ,
meaning that with na ∝ nb ∝ 1/a3 and T ∝ 1/a (for the radiation era, see Section 7.3.1) we get
1
(7.60)
α ∝ 13/2
a
How large must α be to achieve this equilibrium? The trick is to compare τlife = 1/α
with the Hubble time 1/H. So if α % H then we can expect that n s is in equilibrium,
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while if α & H equilibrium is no longer guaranteed and freeze-out sets in. For α , H
we essentially have no more collisions, and the population has frozen out completely,
behaving as n s ∝ 1/a3 . With H ∝ 1/a2 (cf. Eq. 4.50) we get
α
1
∝
H a9/2
(7.61)
This means that as we go toward smaller a we will inevitably find a point where α/H
starts exceeding unity, i.e. where thermal equilibrium is guaranteed. Likewise, for
sufficiently large a there will be a point where collisions can be ignored.
If we define
N s = n s a3
N sT = n sT a3
and
(7.62)
then Eq. (7.59) becomes
%
&
d ln N s
N sT
=α
−1
dt
Ns
(7.63)
%
&
d ln N s
α N sT
=
−1
d ln a
H Ns
(7.64)
With Hdt = d ln a we get
Here again you clearly see the importance of the α/H ratio in determining whether N s
is always forced to be close to N sT or not. If N sT does not change with a, then once
N s & N sT , this will stay like that even after α/H , 1.
If N sT (t) changes with time, then N s (t) will follow N sT (t) as long as α/H % 1. Once
α/H ! 1 the value of N s (t) will flatten-off. This is called “freeze-out”.
As we have seen in Sections 7.3.1 and 7.2.5, radiation has n ∝ 1/a3 , i.e. for radiation
particles we would have N sT (t) =constant. This means that radiation does not really
“freeze out” because even after it is thermally decoupled from the rest of the matter,
N s stays at the thermal value N sT automatically. This is one of the reasons why the
CMB is so extremely close to a perfect blackbody spectrum, as we shall see later.
There is also another type of “freeze-out” that we will often encounter: even if the
reaction timescales are still much smaller than the expansion timescale, the temperature may drop below the threshold for the reaction. This meanins that the reaction will
then, in equilibrium, quickly go in one direction: toward the lower energy state. Under
the conditions we will encounter, where the baryons are emersed in a sea of photons
with much higher number density, this “shut off” can be a quite steep function of
temperature and thus happen rather abruptly.
7.4
Thermal events from 0.1 second until recombination
Now finally we have all the tools in place to study the formation of the elements in
the early Universe, effects such as the production of the cosmic neutrino background
and, finally, the recombination of the Universe and the release of the CMB. But before
doing that we must estimate the number density of baryons.
7.4.1 Baryon to photon ratio
In the very early Universe (, 10−12 seconds after the Big Bang) there was some, as
yet not well understood, mechanism that produced an asymmetry in the number of
Baryons over anti-Baryons. This asymmetry was small:
n-b − n-b
n-b
∼ 10−9
(7.65)
where n-b and n-b are the number densities of baryons and anti-baryons at this early
time. But without this asymmetry the Universe today would not have contained any
65
atoms. It is currently a major field of research in theoretical physics to explain this
asymmetry. The CP-violation (violation of the supposed symmetry of charge and
parity of particles) discovered experimentally in the 60s may have something to do
with it, but the jury is still out.
At some point the Universe had cooled down enough that all baryon-anti-baryon pairs
annihilated, leaving the excess of baryons as the only surviving baryonic matter:
nb = n-b − n-b
(7.66)
Today we can observationally determine nb , but can we know what the original n-b
and n-b were, i.e. how big the asymmetry was? In thermal equilibrium the number
of baryons n-b & n-b must have been comparable to the photon density n-γ according
to Eq. (7.41). It therefore gives an estimate of the asymmetry to compare nb to the
number of photons nγ from the CMB today:
η :=
nb
nγ
(7.67)
Both nb and nγ scale as 1/a3 as we know from Sections 7.3.1 and 7.3.2, that is: as
long as the total particle number does not change. For the baryons that number has
not changed since the annihilation of baryons and anti-baryons. For the photons of the
CMB that number has last changed around 1 second after the Big Bang (see Section
7.4.2). So if we measure η today, it is valid back to 1 second after the Big Bang.
As we shall see later, the baryon density Ωb,0 can be estimated from the anisotropies
of the CMB. Its value is, according to the latest WMAP results: Ωb,0 = 0.0456 (see
Table 4.1), which at the present time, with mb the proton mass mb = m p = 1.67 ×10−24
gram leads to
Ωb,0 ρcrit,0
=& 2.5 × 10−7 (p, n)/cm3
(7.68)
nb =
mp
The number density of photons today is by far dominated by the CMB, for which we
know the temperature exactly T CMB = 2.725 K. With Eq. (7.41) gives
nγ = 10.14gγ(T/K)3 = 410 photons/cm3
(7.69)
η = 6 × 10−10
(7.70)
This gives
This shows (a) how small the baryon-anti-baryon asymmetry was and (b) that for
every baryon in the Universe there are over 1 billion photons! The latter fact plays an
important role in what follows. Another important consequence is that the entropy of
the Universe is dominated by a huge margin by the CMB photons.
7.4.2 Neutrino background and electron-positron sea
Around 0.1 seconds after the Big Bang the Universe was filled with electron-positron
pairs, neutrino-antineutrino pairs and photons, all with their thermal equilibrium abundances. As we saw in Section 7.4.1 there were also neutrons and protons, but their
number densities and total energy density were extremely low compared to those of
the neutrinos, electron/positrons and photons. The density of the Universe was so large
that neutrinos were routinely absorbed and re-created through the weak interaction:
ν + ν ↔ e+ + e−
(7.71)
At temperatures of about 1010.5 K these interactions froze out and the neutrino background was released. It should still be with us today, but it is extremely hard to detect,
so we have no proof of its existence yet.
66
After the freeze-out of the weak interaction, the sea of e+ , e− was kept at thermal
equilibrium through
γ + γ ↔ e+ + e−
(7.72)
As the Universe cooled down, however, there inevitably comes a time when the average photon energy 'hν( = 2.7kT drops below the rest mass energy me c2 of an electron,
which happens at
me c 2
T&
= 2.2 × 109 K
(7.73)
2.7k
After that time the forward direction of reaction Eq. (7.72) becomes more and more
difficult and very rapidly all e+ , e− pairs annihilate through
e+ + e− → γ + γ
(7.74)
Before this happened the temperatures of the photon sea and the neutrino sea remain
the same, as they both cool in the same way (cf. Section 7.3.1). However, when
the electron-positron pairs suddenly vanish, the energy that was initially in both the
pairs and the photons suddenly has to be accounted for by only the photons. The
temperature scaling of Section 7.3.1 was valid if the number of particles (photons in
this case) remains constant in time, or more properly formulated: if the entropy of
the photon gas stays the same. However, reaction Eq. (7.74) shows that the electrons
give all their energy (entropy) to the photons. Once this process was finished the
photons had substantially more entropy than before, and thus would have been hotter
than Section 7.3.1 predicted. During the pair annihilation phase the temperature of
the photon gas therefore decreased only very slowly, as the adiabatic cooling was
continuously compensated by γ + γ productions.
If we write the entropies of e+ , e− and γ before the annihilation phase as s-e+ , s-e− and s-γ
and the entropy of the γ after the annihilation phase as sγ then the transfer of entropy
from the pairs to the photons can be written as
s-e+ + s-e− + s-γ = sγ
(7.75)
The statistical weights of all three particles is 2: ge+ = ge− = gγ = 2. With these
statistical weights we can use the formulae from Section 7.2.5 to infer that
s-e+ = s-e− =
7 s
8 γ
(7.76)
Eq. (7.75) then becomes
%
&
7 7
11 + + 1 s-γ =
s = sγ
8 8
4 γ
Since sγ ∝ T 3 we get
Tγ =
%
11
4
&1/3
T γ- & 1.4T γ-
(7.77)
(7.78)
meaning that the photon gas after annihilation is hotter by a factor of 1.4 compared
to the case if no annihilation would have taken place. Since the neutrinos have not
experienced any of this, we must infer that
T γ & 1.4T ν
(7.79)
This is still so today, so that we expect the cosmic neutrino background temperature to
be around 1.95 K. The energy density of the neutrino background is about 1/4 that of
the CMB, as can be derived from Eq. (7.42). In these derivations we have, however,
assumed that the neutrinos all have neglible masses. If the neutrino masses are nonneglible they may have cooled down much further than this.
67
7.4.3 Freeze-out of neutron-proton ratio
While the baryon number density is fixed by the baryon-anti-baryon asymmetry, their
identity (whether they are neutron or proton) around the time of neutrino freeze-out is
not yet determined. This is because of weak interactions such as
n + νe ↔ p + e−
(7.80)
Well before the annihilation of the positrons the number density of electrons is very
much higher than the number of protons, and so is the number of neutrinos. Both
populations have thermal number densities and hence zero chemical potential. That
means that reaction Eq. (7.80) does not appreciably affect the number densities of neutrinos and electrons, so that in this reaction they are unconstrained. Only the number
of protons and neutrons are constrained. Their thermal abundance ratio is therefore
nn
2
= e−∆mc /kT
np
(7.81)
where ∆m = 1.4 MeV is the mass difference between a neutron and a proton. Reaction
Eq. (7.80) freezes out at temperatures around 800 keV & 9 × 109 K. This gives a
neutron-proton ratio of about
nn
1
&
(7.82)
np 6
at the time of freeze-out.
7.4.4 Formation of deuterium
Around 3 minutes after the Big Bang, the temperature has dropped to about 80 keV
& 9 × 108 K which is low enough for neutrons and protons to fuse and form deuterium.
By this time about 20% of the neutrons have already spontaneously decayed through
the process:
n → p + e− + νe
(7.83)
because the half-life of neutrons is 887 seconds & 15 minutes. So by the time fusion
starts we have
1
nn
&
(7.84)
np 7
The fusion of n and p is given by the reaction
n+ p↔ D+γ
(7.85)
This reaction does not entirely proceed in equilibrium (see later), but one can use equilibrium theory to get a feeling for the temperature at which deuterium can form. We
use the method used to derive the Saha equation for ionization of Hydrogen (Section
7.2.4). Compared to that section we replace e with n and H with D and we arrive at
Nn N p Zn Z p V(2πmn m p kT )3/2 −χ/kT
e
=
=
3
ND
ZD
m3/2
D h
(7.86)
with χ & 2 MeV being the binding energy of deuterium. If we write x = Nn /Nn0 =
Nn /(Nn + ND ), we can replace (using Nn0 /N p0 = 1/7, see Eq. 7.84): ND = Nn0 (1 − x),
N p = Nn0 (x + 6) and we get
x(x + 6) (2πmn m p kT )3/2 −χ/kT
=
e
3 0
1−x
m3/2
D h nn
(7.87)
with n0n = Nn0 /V the initial neutron number density before deuterium formation. Now,
n0n = nb /8 where nb is the baryon number density. We know from Section 7.4.1 that
68
nb = ηnγ with η = 6 ×10−10 . We also know that the photons are in thermal equilibrium
so that with Eq. (7.41) we have nγ = 2(ζ(3)/π2!3 c3 )(kT )3 This means that we get
x(x + 6) 1 (mn m p c2 )3/2
=
e−χ/kT
1−x
η ζ(3)π(2πmDkT )3/2
(7.88)
x = 1 means no deuterium has formed while x = 0 means that all neutrons have been
consumed to form deuterium. Eq. (7.88) contains a factor 1/η % 1. This means that
order for x to become appreciably smaller than 1 the factor χ/kT in the exponent must
be very large. Or in other words: the production of deuterium happens at a temperature
kT , χ. Naively one would expect that this happens around kT & χ & 2 MeV. But
because there are so many photons around that can destroy deuterium, the formation
of deuterium happens only much later at kT & 80 keV.
7.4.5 Formation of helium and lithium
The formation of deuterium is essential for the formation of higher mass elements.
In order to form elements such as 3 He and 4 He directly out of neutrons and protons
one would require 3-body or 4-body collisions to happen. The probabilities for such
reactions are so low that they can be considered zero. With deuterium present, further
elements can be formed through reactions such as:
D+D
3
He + D
D+n
T+ p
↔
↔
3
4
He + n
He + p
↔ T+γ
↔
4
He + γ
(7.89)
(7.90)
(7.91)
(7.92)
where T denotes tritium (3 H). This means that it must be possible to form deuterium
quickly enough, and deuterium must be long-lived enough, otherwise the Universe
would not contain any elements other than hydrogen. If, however, the formation of
deuterium would be too efficient, then all deuterium would have reacted further to
helium, so today there would be no deuterium left. This leads to the Gamov condition,
that
'n p vσ(t & 1
(7.93)
where v is the typical velocity of neutrons and protons, σ is the reaction cross section
for the formation of deuterium, n p is the proton number density and t the time scale of
nucleosynthesis.
In the end, most of the neutrons end up in 4 He. It is, however, not easy to form higher
mass elements. The reaction
T +4 He ↔7 Li + γ
(7.94)
causes the formation of Lithium in a tiny amount. Also trace amounts of Beryllium is
formed, which is, however, unstable and decays. The main barrier to forming larger
mass elements is the absense of stable nuclei of atomic weights 5 and 8. It is therefore
not possible to produce a larger mass nucleus from 4 He by capturing a neutron or
proton, and likewise for 7 Li. This means that beyond 7 Li essentially no elements are
formed. All the heavier elements in the Universe today have been formed in stars.
Since most of the neutrons end up in 4 He we can estimate the helium abundance
straight from the n/p = 1/7 ratio:
mHe nHe
4(nn /2)
2nn
=
=
mHe nHe + mH nH
4(nn /2) + (n p − nn ) nn + n p
2(nn /n p )
1
=
=
1 + (nn /n p ) 4
Y=
(7.95)
This is a rather robust prediction, and fortunately this is indeed very close to what is
observed!
69
7.4.6 Observational constraints from nucleosynthesis
The primordial abundances are not easy to measure, because most of the matter in the
Universe has already been processed into further elements by stellar nucleosynthesis.
Therefore one must either observe high redshift objects or find ultra-low metallicity
stars in our own Galaxy. But even then it is not easy, because, for instance, deuterium
is electronically identical to hydrogen. If deuterium is in the form of a molecule such
as DH, then the ro-vibrational spectrum has its lines shifted compared to H2 because
of the intertia of D. Also a tiny shift in the location of the Ly-α line in neutral atomic
deuterium can be observed. Likewise the abundance of 3 He is not easy to determine,
but can be detected via the hyperfine transition of 3 He+ . However, 3 He is not a very
reliable tracer because it can be created in pre-main-sequence stars and destroyed in
stellar interiors. 4 He is easier to detect: via optical recombination lines in HII regions.
7
Li can be found in extremely metal-poor star in the galactic halo.
In the lecture, some examples will be shown.
7.4.7 Recombination of the Universe and the release of the CMB
The CMB is released when the electrons, which scatter the radiation and thus make
the Universe opaque, recombine with the available ions, or in other words, when the
reaction
e− + p ↔ H + γ
(7.96)
freezes out. This happens when the temperature of the Universe drops well below the
ionization energy. Like with the n + p ↔ D + γ reaction in Section 7.4.4 this is a
delayed recombination: it happens at a temperature of about 3500 K = 0.3 eV, which
is substantially lower than the ionization energy of hydrogen (13.7 eV). The reason is,
again, the tiny baron-to-photon ratio η. The Saha equation (Eq. 7.36) can be written,
with nb = ηnγ as
(2πme kT )3/2 −χ/kT
x2
=
e
(7.97)
1−x
h3 ηnγ
with χ = 13.7 eV. With nγ = 2(ζ(3)/π2!3 c3 )(kT )3 (cf. Eq. 7.41) this becomes
%
&3/2
%
&3/2
√
x2
π
me c 2
0.26 me c2
−χ/kT
= √
e
&
e−χ/kT
1 − x 4 2ζ(3)η kT
η
kT
(7.98)
Like with the deuterium production, the huge factor 1/η means that in order to get
recombination (x → 0) the factor χ/kT must be % 1, delaying recombination until
3500 K.
Also, like with the deuterium production, the assumption of strict thermal equilibrium
at all times is here a bit too simple. One problem is, for instance, the Ly-α transition at
10.2 eV. It is the 2 P↔2 S electronic transition from the first excited state to the ground
state. While such photons cannot directly ionize other hydrogen atoms, they will keep
many atoms in a highly excited state (just 3.5 eV below the ionization threshold),
making it very easy to reionize them. Moreover, Ly-α photons cannot escape: any
photon that is emitted by one hydrogen atom will be reabsorbed by a nearby H atom
in the ground state, as long as they are around.This leads to Ly-α resonant scattering:
When a hydrogen atom absorbs a Ly-α photon it stays excited until it deexcites by
sending out a Ly-α photon, which moves a little distance before it gets absorbed again.
This process keeps going on as long as there is no mechanism to deexcite the 2 P
and 2 S states of hydrogen in another way. One way would be through a collision
with another hydrogen atom, but this rate of collisional de-excitation is rather low.
Another way would be by two-photon deexcitation, which is a quadrupole transition
and thus “highly forbidden”, meaning that its rate is very low compared to “normal”
electronic transitions - but still higher than collisional de-excitation at the conditions
in the Universe around 3500 K. Since this quadrupole process produces two photons,
70
the energetic constraint is hν1 + hν2 =10.2 eV. This gives a continuum of photons that
are less energetic than Ly-α. These photons do not excite other atoms, meaning that
the Ly-α scattering process ends and one atom is now recombined “for real”. The low
rate of quadrupole emission means that recombination is delayed again a bit.
The recombination process eliminates free electrons and thus eliminates the Thompson scattering opacity of the gas. Neutral hydrogen+helium gas is almost perfectly
transparent for optical radiation. This means that the radiation is now set free: the
CMB is born. The point of release is rather abrupt, but not infinitely abrupt. There is
a certain width in time where the Universe goes from completely opaque to the time
when a photon scatters for the last time (the “last scattering surface”). The optical
depth through this surface to Thompson scattering is given by
#
#
τ=
ne σT dr = nB σT
xdr
(7.99)
with σT = 6.65 × 10−25 cm2 the Thompson cross section and dr = cdt = cda/ȧ. The
chance that a photon scatters for the last time between z and z + dz is
p(z)dz = e−τ dτ = e−τ
dτ
dz
dz
It turns out that a reasonably good description of p(z) is
&
%
1
(z − zrec )2
p(z) & √
exp −
2σ2rec
2πσrec
(7.100)
(7.101)
with zrec = 1100 and σrec = 80. This means that the last scatterings happen over a time
range during which the temperature drops by roughly 200 K (with 3500 K average).
This does not mean that the CMB has a temperature uncertainty of (200/3500) ∗
2.725 = 0.156 K, because the temperature of the free photons scales the same as those
of the trapped photons. Note that although we are by now in the matter-dominated
era, the energy of matter is dominated by its rest mass, so that the thermal energy
is still dominated by the radiation field, which means that the temperature scaling is
T ∼ 1/a, meaning that we obtain a CMB of a single temperature even though the light
is emitted over a range of temperatures.
Once the CMB radiation is set free, the radiation can no longer influence the baryon
temperature. Before the release of the CMB the baryon temperature was always kept
at the radiation temperature, thus also going as T ∝ 1/a. But after the CMB release
the baryons cool as a monoatomic gas: T ∝ 1/a2 .
71